Darboux's Theorem

Let $f$ be differentiable on a closed interval $I = [a, b]$ and let $α \in R$ satisfy either that:

$f^{'} (a) < α < f^{'} (b)$ , or
$f^{'} (a) > α > f^{'} (b)$
Then $\exists c \in (a, b)$ such that $f^{'} (c) = α$ .

This uses the Derivative definition so make sure to know that beforehand.

Proof

The trick here is to use the Interior Extremum Theorem, over and over again. We'll have issue using it though because it works for an open interval, and not a closed interval like the Extreme Value Theorem does.

Let $g (x) = f (x) - α x$ . By the Algebraic Differentiability Theorem then $g$ is differentiable on $[a, b]$ and $\forall x \in [a, b]$ then:

g^{'} (x) = f^{'} (x) - α

Now for the first case, suppose that $f^{'} (a) < α < f^{'} (b)$ . Then:

g^{'} (a) = f^{'} (a) - α < 0

g^{'} (b) = f^{'} (b) - α > 0

So then $g^{'} (a) < 0 < g^{'} (b)$ . Since $g$ is differentiable on $[a, b]$ , then it is continuous on that interval. Using the Extreme Value Theorem then $g$ has an absolute minimum on $[a, b]$ , denote it $c$ .

We want $c \in (a, b)$ so it goes to show that $c \neq a, b$ . Notice that at $a$ :

0 > g^{'} (a) = lim_{x \to a^{+}} \frac{g (x) - g (a)}{x - a}

Now notice that $x - a > 0$ for this limit, and thus that suggests that $g (x) - g (a) < 0$ by the Limits and Order (Order Limit Theorem). So then $\exists x_{0} > a$ such that $g (x_{0}) - g (a) < 0$ , so then we found a smaller value than $g (a)$ ! This contradicts $a$ being our supposed minimum.

Similarly for $b$ :

0 < g^{'} (b) = lim_{x \to b^{-}} \frac{g (x) - g (b)}{x - b}

Now $x - b < 0$ and thus by the Limits and Order (Order Limit Theorem), then $\exists x_{1} < b$ such that $g (x_{1}) - g (b) < 0$ so then $b$ cannot be a minimum!

Thus $c \in (a, b)$ , and by the Interior Extremum Theorem then $g^{'} (c) = 0$ . Correspondingly then $f^{'} (c) = α$ .

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